Roll a die $n$ times ($n$ is a natural number). What is the probability that 1 and 6 are observed at least once? Note: This is a homework problem so I cannot accept solutions. I would like suggestions as to how to proceed. 
I have that each trial of rolling a die is independent. So I can say:   
Let $P(1)$ = Rolling a 1 and $P(6)$ = Rolling a 6. We want to find $P(1\cap 6)$. Since rolling a die and getting an outcome in a trial is independent of other trials: $P(1\cap6)=P(1)P(6) $
It's unclear to me how to find the probability that you roll a 1 if you roll $n$ times. Would it be $P(1)=1/6^n$
 A: The total number of combinations is $6^n$.
The number of combinations which include only $[2,3,4,5,6]$ is $5^n$.
The number of combinations which include only $[1,2,3,4,5]$ is $5^n$.
The number of combinations which include only $[2,3,4,5]$ is $4^n$.

So the probability of not observing $1$ and $6$ is therefore:
$$\frac{5^n+5^n-4^n}{6^n}$$
And the probability of observing $1$ and $6$ at least once is:
$$1-\frac{5^n+5^n-4^n}{6^n}$$
A: Let $t_{ab}(n)$ be the number of sequences of length $n$ containing $a$ and $b$ (assuming $a \ne b$).
Let $t_{a}(n)$ be the number of sequences of length $n$ containing $a$.
Let $t(n)$ be the number of sequences of length $n$.
The question is to investigate $t_{16}(n)$.  Each sequence starts with a $1$, $6$, or something else, so:
$$t_{16}(n) = t_{1}(n - 1) + t_{6}(n - 1) + 4 ~ t_{16}(n-1) \tag{A}$$
Similarly, $t_{a}(n)$ either starts with $a$ or it doesn't:
$$t_{a}(n) = t(n-1) + 5 ~ t_{a}(n - 1) \tag{B}$$
Finally, 
$$t(n) = 6^n \tag{C}$$
(C) is solved, so (B) becomes:
$$t_{a}(n) = 6^{n-1} + 5 ~ t_{a}(n - 1) \tag{B2}$$
A recursive equation which can be solved:
$$\begin{align}
%
 -1~t_{a}(n) + 5~t_{a}(n-1) &= 6^{n-1} \\
%
 -1~t_{a}(n + 1) + 5~t_{a}(n) &= 6\cdot 6^{n-1} \\
%
t_{a}(n + 1) &= 11~t_{a}(n) - 30~t_{a}(n-1) \\
%
\end{align}$$
Which is a linear recursive equation, and it solves to
$$t_{a}(n) = 6^n - 5^n \tag{B3}$$
So (A) becomes
$$t_{16}(n) = 2\left(6^{n-1} - 5^{n-1}\right) + 4 ~ t_{16}(n-1) \tag{A2}$$
Solving similarly:
$$\begin{align}
%
t_{16}(n) - 4 ~ t_{16}(n-1) &= 2~6^{n-1} - 2~5^{n-1} \\
%
t_{16}(n+1) - 4 ~ t_{16}(n) &= 12~6^{n-1} - 10~5^{n-1} \\
%
t_{16}(n+2) - 4 ~ t_{16}(n+1) &= 72~6^{n-1} - 50~5^{n-1} \\
\end{align}$$
Eliminating the exponents between the 3 equations, it yields:
$$t_{16}(n+2) = 15~t_{16}(n+1) - 74~t_{16}(n) + 120~t_{16}(n-1)$$
Another linear recursive equation, which solves to:
$$t_{16}(n) = 6^{n} - 2\cdot 5^{n} + 4^{n} \tag{A3}$$
And the resulting probability is then:
$$\boxed{\huge{\frac{6^{n} - 2\cdot 5^{n} + 4^{n}}{6^n}}}$$
